13. Equations with absolute values
Earlier we learned |±a|=a
|a|, also called the "absolute value" of a, returns the value of a after removing its sign.
So, how do we now solve an equation of the form \$|x+3|=5\$?
One way would be to square both sides, since squaring also effectively removes a negative sign.
(squares of negative and positive numbers are both positive)
This gives: \$|x+3|^2=5^2\$ or \$x^2+6x+9=25\$ or \$x^2+6x-16=0\$
\$x^2+6x-16=(x+8)(x-2)=0\$, from which we get \$x=-8\$ or \$x=2\$
Substituting, we see \$|-8+3|=|-5|=5\$ and \$|2+3|=|5|=5\$

e.g. solve: \$|x+3|<5\$
Squaring both sides, \$|x+3|^2<5^2\$ or \$x^2+6x+9<25\$ or \$x^2+6x-16<0\$
This gives, \$x^2+6x-16=(x+8)(x-2)<0\$,
which means only one of \$(x+8)\$ or \$(x-2)\$ is less than 0 (for their product to be negative)
if \$(x+8)<0\$, then \$x<-8\$, and \$x-2\$ is also negative, which wouldn't work.
so \$(x-2)<0\$, in which case \$x<2\$ and \$(x+8)>0\$ or \$x>-8\$, which gives us a range of \$-8<x<2\$ as the desired solution.